Rigid Body Motion with Fixed Point Suppose that $  T : \mathbb{R}  \longmapsto    \mathbb  SO_{3}  $ is a differentiable function. This means that $ T(t) $ is differentiable component wise. Does there exist a function $     \Omega  :  \mathbb{R}  \longmapsto     \mathbb{R}^{3} $ such that $ T'(t)  \vec{x}  =     \Omega(t)   \times  T  \vec{x}  $ for all $  \vec{x}  \in   \mathbb{R} ^{3}  $ and for all $ t \in \mathbb{R} $.      
If so, is $   \Omega $ unique? Can $ \Omega $ be chosen to be differentiable? 
Edit 1: changed $   \vec{x} $ to $ T \vec{x} $. Thanks to the example below, I think the correct formulation is what I have written above. 
Edit 2: Some progress:   
We know that $  T(t)  T^{T}(t) =  I $, and thus, $  T'(t)  T(t)  ^{T}    +  T(t)    T'(t) ^{T}  =  0    $. This implies that $  T'(t) T(t  )   ^   {   T   }          $ is skew-symmetric, which means that $ \det (  T'(t) T ^{T}(t)  ) =  0    $. This implies that $  T'(t)    $ is singular, which means that  $ \ker (  T'(t)   ) $ is non-trivial.  
Also, we have  $ T(t)   \vec{x}   \cdot  T(t)  \vec{x} =     \vec{x}   \cdot   \vec{x}     \implies  2  T (t) \vec{x}      \cdot   T ' (t)   \vec{x}  = 0     $.   
 A: Yes, this $\Omega$ exists, is unique, and is called the Darboux vector.
Rotations form a Lie group; in particular, you can write $T(t_0+\delta t) = R(\delta_t, t_0) T(t_0)$ where $R$ is a second rotation that you can think of as a small correction to $T(t_0)$, with $\lim_{\delta t\to 0} R = I$. The precise formula for $R$ is a bit involved, due to the non-commuting nature of a rotations, but it can be shown that the infinitesimal rotation $\frac{\partial R}{\partial \delta t}\vert_{\delta t\to 0}$ is a skew-symmetric matrix (which you can write as $\Omega(t_0) \times$ for some vector $\Omega(t_0)$).
If you need formulas for computation, here is a very helpful paper I dug up when I was recently preparing for a lecture on rigid body dynamics: https://smartech.gatech.edu/bitstream/handle/1853/52144/DerivativeOfRotation.pdf?sequence=1
A: I think that in general we don't have such a function $\Omega$. Consider for example
$$
T(t) = \begin{pmatrix}\cos t & -\sin t & 0 \\ \sin t & \cos t & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
Then
$$
T'(t) = \begin{pmatrix}-\sin t & -\cos t & 0 \\ \cos t & -\sin t & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
Now pick
$$
x = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}.$$
Then we have
$$
T'(t)x = \begin{pmatrix}-\sin t \\ \cos t \\ 0 \end{pmatrix}$$
and in particular
$$
T'(\pi/4)x = \begin{pmatrix}-\frac1{\sqrt2} \\ \frac1{\sqrt2} \\ 0 \end{pmatrix}.$$
This is not orthogonal to $x$. But $\Omega(\pi/4) \times x$ is orthogonal to $x$ for any $\Omega(\pi/4)$. So $\Omega$ can't exist.
